Results 1 to 3 of 3

Math Help - can someone check my work? geometric series

  1. #1
    Newbie
    Joined
    Oct 2006
    Posts
    9

    can someone check my work? geometric series

    okay I attached the two questions and I will show my work here
    question#1
    sn= a(1-r^n) / (1-r)

    a = 16
    r = 0.5
    n = 50
    so:
    sn= a(1-r^n)/ (1-r)
    sn= 16(1-(0.5)^50)/ (1-(0.5))
    sn= 16(1-(0.5)^50)/ (0.5)
    sn= 32


    question#2
    sn= a(1-r^n)/ (1-r)
    a = 3*(1-.07)
    r =(1-0.07)
    n = 7
    So..
    sn= a(1-r^n)/ (1-r)
    sn= 3(1.07)x(1-(1-0.07)^7)/ (1-(1-0.07))
    sn= 3(1.07)x (1-(1-0.07)^7)/(.07)
    sn= 1.83 I rounded to two decimal places!

    thanks for checking!
    Last edited by frogsrcool; June 8th 2007 at 10:05 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,962
    Thanks
    349
    Awards
    1
    Quote Originally Posted by frogsrcool View Post
    okay I attached the two questions and I will show my work here
    question#1
    sn= a(1-r^n) / (1-r)

    a = 16
    r = 0.5
    n = 50
    so:
    sn= a(1-r^n)/ (1-r)
    sn= 16(1-(0.5)^50)/ (1-(0.5))
    sn= 16(1-(0.5)^50)/ (0.5)
    sn= 32


    question#2
    sn= a(1-r^n)/ (1-r)
    a = 3*(1-.07)
    r =(1-0.07)
    n = 7
    So..
    sn= a(1-r^n)/ (1-r)
    sn= 3(1.07)x(1-(1-0.07)^7)/ (1-(1-0.07))
    sn= 3(1.07)x (1-(1-0.07)^7)/(.07)
    sn= 1.83 I rounded to two decimal places!

    thanks for checking!
    The first one is correct. The setup for the second one is correct, but you have a math error somewhere. I can point out one thing: Your a = 1-0.07 = 0.93, not 1.07. Anyway, I got the sum to be about 15.8750653.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,749
    Thanks
    650
    Hello, frogsrcool!

    1)\;\sum^{50}_{n=1} 8(0.5)^{n-2}

    S_n\:= \:a\frac{1 - r^n}{1 - r}

    a = 16,\;r = 0.5,\;n = 50

    so: . S_{50}\:=\:16\frac{1 - 0.5^{50}}{1 - 0.5}

    Therefore: . S_{50} \:\approx\:32

    Since (0.5)^{50} \:\approx\:8.88 \times 10^{-16}, your answer is close enough!



    2)\;\sum^7_{n=1}3(1-0.07)^n

    a \;= \;3(1 - 0.07) \;=\;3(0.93) \;=\;2.79
    r \;= \; 1 - 0.07 \;=\;0.93
    n \;= \;7

    S_7 \;=\;2.79\,\frac{1 - 0.93^7}{1 - 0.93} \;=\;2.79\,\frac{0.398299129}{0.07} \;=\;15.8750653

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Check my work on alternating series problem?
    Posted in the Calculus Forum
    Replies: 2
    Last Post: April 29th 2010, 04:31 PM
  2. Check my work on Taylor series?
    Posted in the Calculus Forum
    Replies: 8
    Last Post: May 11th 2009, 07:37 PM
  3. Maclaurin series - check my work
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 19th 2009, 06:07 PM
  4. Binomial series - check my work?
    Posted in the Calculus Forum
    Replies: 2
    Last Post: April 16th 2009, 08:36 PM
  5. Can someone check this geometric series??
    Posted in the Algebra Forum
    Replies: 1
    Last Post: March 7th 2007, 06:11 PM

Search Tags


/mathhelpforum @mathhelpforum